Question

Find the equation of a plane which passes through \[\left( {2, - 3,1} \right)\] and is normal to the line joining the points \[\left( {3,4, - 1} \right)\] and \[\left( {2, - 1,5} \right)\].
A. \[x + 5y - 6z + 19 = 0\]
B. \[x - 5y + 6z - 19 = 0\]
C. \[x + 5y + 6z + 19 = 0\]
D. \[x - 5y - 6z - 19 = 0\]

Answer 1

Hint: First, use the formula of the plane passing through a point and calculate the equation of the required plane. Then, calculate the direction ratios of the line joining the points \[\left( {3,4, - 1} \right)\] and \[\left( {2, - 1,5} \right)\]. After that, substitute the direction ratios in the equation of plane and solve it to get the required answer.

Formula used: The equation of plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\] is: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\] .
The direction ratios of the line joining the points \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\] is: \[\left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)\]

Complete step by step solution: Given:
A plane passes through the point \[\left( {2, - 3,1} \right)\].
Plane is normal to the line joining the points \[\left( {3,4, - 1} \right)\] and \[\left( {2, - 1,5} \right)\].

Let’s calculate the equation of the required plane.
Apply the formula of the equation of plane passing through the point.
We get,
The equation of the required plane with the direction ratios \[\left( {a,b,c} \right)\] is:
\[a\left( {x - 2} \right) + b\left( {y + 3} \right) + c\left( {z - 1} \right) = 0\] \[.....\left( 1 \right)\]

The given line joins the two points \[\left( {3,4, - 1} \right)\] and \[\left( {2, - 1,5} \right)\].
Apply the formula of the direction ratios of a line.
We get,
Direction ratios: \[\left( {2 - 3, - 1 - 4,5 + 1} \right)\]
\[ \Rightarrow \] Direction ratios: \[\left( { - 1, - 5,6} \right)\]

It is given that the plane is normal to the line joining the points \[\left( {3,4, - 1} \right)\] and \[\left( {2, - 1,5} \right)\].
So, the direction ratios of the plane and the line are equal.
We get, \[a = - 1,b = - 5,c = 6\]
Now substitute these values in the equation \[\left( 1 \right)\].
 \[ - 1\left( {x - 2} \right) - 5\left( {y + 3} \right) + 6\left( {z - 1} \right) = 0\]
\[ \Rightarrow - x + 2 - 5y - 15 + 6z - 6 = 0\]
\[ \Rightarrow - x - 5y + 6z - 19 = 0\]
\[ \Rightarrow x + 5y - 6z + 19 = 0\]
Thus, the equation of the plane which passes through \[\left( {2, - 3,1} \right)\] and is normal to the line joining the points \[\left( {3,4, - 1} \right)\] and \[\left( {2, - 1,5} \right)\] is \[x + 5y - 6z + 19 = 0\].

Thus, Option (A) is correct.

Note: Students often get confused between the direction ratios and direction cosines. Remember the following formulas:
For the line joining the points \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\]:
Direction ratios: \[\left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)\]
Direction cosines: \[\left( {\dfrac{{{x_2} - {x_1}}}{{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} }},\dfrac{{{y_2} - {y_1}}}{{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} }},\dfrac{{{z_2} - {z_1}}}{{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} }}} \right)\]